The formula for the sine sum is
[tex]\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)[/tex]
So, equation
[tex]\sin(a)\cos(b)+\cos(a)\sin(b) = \sin(a)+\sin(b)[/tex]
Cannot be solved with standard ways. However, we can build special cases to make it work.
The most trivial example would be A=B=0. In this case, we have
[tex]\sin(0+0)=0 \sin(0) = 0,\quad\sin(0)+\sin(0)=0+0=0[/tex]
If we work on this idea a little bit, we can produce examples like
[tex]A=2k_1\pi,\quad B=2k_2\pi[/tex]
And again we have
[tex]\sin(A+B)=\sin(2\pi(k_1+k_2))=0[/tex]
[tex]\sin(A)+\sin(B)=\sin(2k_1\pi)+\sin(2k_2\pi)=0+0=0[/tex]
So, yes, it is possible that [tex]\sin(A+B)=\sin(A)+\sin(B)[/tex], although solving the equation directly would be quite difficult.
Answer: yes, when A & B are on the quadrantals (0°, 90°, 180°, 270°) which results in a sum of 0
Step-by-step explanation:
sin (A + B) = cos A · sin B + cos B · sin A
If cos A = 1 and cos B = 1, then you will end up with sin A + sin B.
The only angle where cos is equal to 1 is at 0°. Notice that sin is equal to 0.
It follows that when cos is equal to -1, sin is also 0. This occurs at 180°
The other option is when cos is equal to 0. This occurs at 90° and 270°.
All of these options result in sin A + sin B = 0
So, the only possibility when sin (A + B) = sin A + sin B is when it equals zero.
